Question

Calculate the free enegry change when `1mol` of `NaCI` is dissolved in water at `298K`. Given:
a. Lattice enegry of `NaCI =- 778 kJ mol^(-1)`
b. Hydration energy of `NaCI - 774.3 kJ mol^(-1)`
c. Entropy change at `298 K = 43 J mol^(-1)`

Answer 1

`Delta_(sol)G^(Theta) = ?`
`Delta_(sol)H^(Theta) = Delta_(ion)H^(Theta) +Delta_(hyd)H^(Theta)`
`Na^(o+)(g) +CI^(Theta) (g) rarr NaCI(s),DeltaH^(Theta) =- 778 kJ mol^(-1)`
or
`NaCI(s) rarr Na^(o+)(g)+CI^(Theta)(g) DeltaH_(1) = 778 kJ mol^(-1) ..(i)`
`Na^(o+)(g)+CI^(Theta) (g) +(aq)rarr NaCI(aq)` or
`Na^(o+)(aq) +CI^(Theta)(aq)`
`DeltaH_(2) =- 774.3 kJ mol^(-1) ....(ii)`
To calculate `Delta_(sol)H^(Theta)`
`NaCI(s) +aq rarr Na^(o+)(aq) +CI^(Theta)(aq)`
`DeltaH^(Theta) = DeltaH_(1) +DeltaH_(2)`
`= 778 - 774.3 = 3.7kJ mol^(-1) = 3700 J mol^(-1)`
`Delta_(sol)S^(Theta) = 43 J mol^(-1)`
`Delta_(sol)G^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta) = 3700 - 298 xx 43 = - 9114J`
`Delta_(sol)G^(Theta) =- 9.114 kJ`